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JEE Main Math Class 12 Differential Equations Part 2 Questions
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© examsnet.com
Question : 7
Total: 124
Let
y
=
y
(
x
)
be the solution of the differential equation
d
y
d
x
+
2
x
(
1
+
x
2
)
2
y
=
x
e
1
(
1
+
x
2
)
;
y
(
0
)
=
0
. Then the area enclosed by the curve
f
(
x
)
=
y
(
x
)
e
1
(
1
+
x
2
)
and the line
y
−
x
=
4
is ______.
[5 Apr 2024 Shift 2]
Your Answer:
Validate
Solution:
👈: Video Solution
IF
=
e
∫
2
x
(
1
+
x
2
)
2
dx
=
e
−
1
1
+
x
2
y
⋅
e
−
1
1
+
x
2
=
∫
x
⋅
e
1
1
+
x
2
⋅
e
−
1
1
+
x
2
dx
y
⋅
e
−
1
1
+
x
2
=
x
2
2
+
c
(
0
,
0
)
⇒
C
=
0
y
(
x
)
=
x
2
2
e
1
1
+
x
2
f
(
x
)
=
x
2
2
A
=
4
∫
−
2
(
x
+
4
)
−
x
2
2
dx
=
18
© examsnet.com
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